Section Formula and Midpoint

The Section Formula (Internal Division)

Suppose a point $P$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$ (so $AP : PB = m : n$). Then

$$P = \left( \frac{m x_2 + n x_1}{m + n},\; \frac{m y_2 + n y_1}{m + n} \right)$$

Key Point: The ratio number nearest to a point multiplies the far coordinate: $m$ (next to $A$ in $AP$) multiplies $x_2$, and $n$ multiplies $x_1$. Memorise the pattern $\dfrac{m x_2 + n x_1}{m+n}$.

[Board Important] Write the ratio as $m : n$ in the same order as the segment $A$ to $B$, then substitute carefully.

The Midpoint Formula

The midpoint of $A(x_1, y_1)$ and $B(x_2, y_2)$ is the special case $m : n = 1 : 1$:

$$M = \left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2} \right)$$

Think of it this way: the midpoint is just the average of the two x-coordinates and the average of the two y-coordinates.

Key Point: Midpoint = average of coordinates. This is one of the most-used formulas in the whole chapter.

[Board Important] The diagonals of a parallelogram bisect each other — so their midpoints coincide. This fact solves many 'find the fourth vertex' problems.

Finding the Ratio of Division

If we know that a point $P(x, y)$ lies on segment $AB$ but not the ratio, let the ratio be $k : 1$ and use the section formula on one coordinate:

$$x = \frac{k x_2 + x_1}{k + 1}$$

Solve for $k$. A positive $k$ means internal division.

This is exactly how we find the ratio in which the x-axis ($y = 0$) or y-axis ($x = 0$) divides a segment: set the relevant coordinate to 0 and solve for $k$.

Key Point: Use $k : 1$ to reduce two unknowns ($m, n$) to one ($k$). Then a positive $k$ confirms internal division.

[JEE Tip] A negative $k$ would indicate external division — outside the scope of the Class 10 board syllabus but useful to recognise.

Points of Trisection

The two points of trisection of a segment divide it into three equal parts. The point closer to $A$ divides $AB$ in the ratio $1 : 2$, and the point closer to $B$ in the ratio $2 : 1$.

Apply the section formula with these ratios to get both points.

Key Point: Trisection points use ratios $1:2$ and $2:1$ — a very common board question.

[Board Important] Don't confuse trisection (three equal parts, two interior points) with bisection (two equal parts, one midpoint).

Solved Examples

Example 1: Midpoint

Find the midpoint of $A(2, 3)$ and $B(8, 11)$.

Solution:

1. $M = \left( \dfrac{2+8}{2}, \dfrac{3+11}{2} \right) = (5, 7)$.

Final Answer: $(5, 7)$.

Takeaway: Midpoint is the average of the coordinates.

Example 2: Section formula

Find the point dividing $A(1, 2)$ and $B(6, 7)$ internally in the ratio $2 : 3$.

Solution:

1. $x = \dfrac{2(6) + 3(1)}{2+3} = \dfrac{12 + 3}{5} = 3$.
2. $y = \dfrac{2(7) + 3(2)}{5} = \dfrac{14 + 6}{5} = 4$.

Final Answer: $(3, 4)$.

Takeaway: $m = 2$ multiplies $B$'s coordinates; $n = 3$ multiplies $A$'s.

Example 3: Ratio in which x-axis divides

In what ratio does the x-axis divide the segment joining $A(1, -3)$ and $B(4, 5)$?

Solution:

1. Let the ratio be $k : 1$; on the x-axis $y = 0$.
2. $0 = \dfrac{k(5) + 1(-3)}{k+1} \Rightarrow 5k - 3 = 0 \Rightarrow k = \dfrac{3}{5}$.
3. Ratio $= 3 : 5$ (positive ⇒ internal).

Final Answer: $3 : 5$.

Takeaway: Set $y = 0$ and solve for $k$ to find where the x-axis cuts the segment.

Example 4: Fourth vertex of a parallelogram

Three vertices of a parallelogram $ABCD$ are $A(1, 2)$, $B(4, 3)$, $C(6, 6)$. Find $D$.

Solution:

1. Diagonals bisect each other ⇒ midpoint of $AC$ = midpoint of $BD$.
2. Midpoint of $AC = \left( \dfrac{1+6}{2}, \dfrac{2+6}{2} \right) = (3.5, 4)$.
3. Let $D = (x, y)$: $\dfrac{4+x}{2} = 3.5$ and $\dfrac{3+y}{2} = 4$.
4. $x = 3$, $y = 5$.

Final Answer: $D = (3, 5)$.

Takeaway: Equate diagonal midpoints to find a missing vertex.

Example 5: Points of trisection

Find the points of trisection of the segment joining $A(2, -2)$ and $B(-7, 4)$.

Solution:

1. Point $P$ ($1:2$): $\left( \dfrac{1(-7)+2(2)}{3}, \dfrac{1(4)+2(-2)}{3} \right) = (-1, 0)$.
2. Point $Q$ ($2:1$): $\left( \dfrac{2(-7)+1(2)}{3}, \dfrac{2(4)+1(-2)}{3} \right) = (-4, 2)$.

Final Answer: $(-1, 0)$ and $(-4, 2)$.

Takeaway: Trisection uses ratios $1:2$ and $2:1$.

Example 6: Midpoint given, find endpoint

The midpoint of $A(3, 4)$ and $B$ is $(5, 7)$. Find $B$.

Solution:

1. $\dfrac{3 + x}{2} = 5 \Rightarrow x = 7$.
2. $\dfrac{4 + y}{2} = 7 \Rightarrow y = 10$.

Final Answer: $B = (7, 10)$.

Takeaway: Use the midpoint equations to back out a missing endpoint.

Example 7: Ratio by y-axis

In what ratio does the y-axis divide the segment joining $A(-3, 1)$ and $B(5, 6)$?

Solution:

1. On the y-axis $x = 0$; let ratio $k : 1$.
2. $0 = \dfrac{k(5) + 1(-3)}{k+1} \Rightarrow 5k - 3 = 0 \Rightarrow k = \dfrac{3}{5}$.
3. Ratio $= 3 : 5$.

Final Answer: $3 : 5$.

Takeaway: y-axis ⇒ set $x = 0$ and solve for $k$.

Example 8: Verify a midpoint

Is $(0, 0)$ the midpoint of $A(-3, -4)$ and $B(3, 4)$?

Solution:

1. Midpoint $= \left( \dfrac{-3+3}{2}, \dfrac{-4+4}{2} \right) = (0, 0)$.

Final Answer: Yes, $(0, 0)$ is the midpoint.

Takeaway: Opposite points about the origin have the origin as midpoint.

Example 9: Section with given ratio 3:1

Find the point dividing $A(-1, 7)$ and $B(4, -3)$ in the ratio $3 : 1$.

Solution:

1. $x = \dfrac{3(4) + 1(-1)}{4} = \dfrac{12 - 1}{4} = \dfrac{11}{4}$.
2. $y = \dfrac{3(-3) + 1(7)}{4} = \dfrac{-9 + 7}{4} = -\dfrac{1}{2}$.

Final Answer: $\left( \dfrac{11}{4}, -\dfrac{1}{2} \right)$.

Takeaway: Substitute the ratio directly; fractional answers are fine.

Example 10: Centre of a diameter

The endpoints of a diameter of a circle are $(-2, 3)$ and $(6, -5)$. Find the centre.

Solution:

1. The centre is the midpoint of the diameter.
2. $\left( \dfrac{-2+6}{2}, \dfrac{3-5}{2} \right) = (2, -1)$.

Final Answer: $(2, -1)$.

Takeaway: The centre of a circle is the midpoint of any diameter.